Concepts of Plastic Analysis

LECTURE 8

Concepts of Plastic Analysis

• Plastic design utilizes the ductility of steel.

Steel grades for plastic design grade Thickness

less than or equal to

Design strength, py, N/mm2

Minimum yield stress, Ys

Minimum ultimate strength, Us

Minimum elongation on gauge length of 5.65√S0

43 16 275 275 410 20-22%

40 265 265 410

63 255 255 410

100 245 245 410

50 16 355 355 490 20-22%

40 345 345 490

63 335 335 490

100 325 325 490

55 16 450 450 550 19%

40 430 430 550

63 415 430 550

100 400 415 550

• In BS5950 , The design strength py should be taken as 1.0Ys but not greater than Us /1.2 =0.833..USwhere Ys and Us are respectively the minimum yield strength ReH and the minimum tensile strength Rm

Requirements of steel for plastic design

• Stress strain diagram has a plateau at the yield stress extending for at least 6 times the yield strain

• Ratio of the specified minimum ultimate tensile strength to specified minimum yield strength is not less than 1.2

• The elongation on a gauge length of 5.65√S0 is not less than 15% where S0 is the original cross sectional area of the gauge length

Structure must be robust to achieve a mechanism of collapse without instability

• Local instability: limits given in table 11

– For rolled section b/T≤ 9ε

– For built up welded section: b/T≤ 8ε

– For webs generally :d/t≤

• Lateral instability

• Frame instability

Basis of plastic design

• Structures are assumed to collapse by the formation of sufficient plastic hinges to create a collapse mechanism

• Fundamental problems of plastic design

– Prediction of the correct collapse mechanism

– Determination of load factor at collapse

– Determination of BMD at collapse

Fundamental requirements for different methods of solving problems • Equilibrium condition: the bending moments must

represent a state of equilibrium between internal forces in the structure and applied loads

• Mechanism condition: At collapse, the bending moment must be equal to the full plastic moment of resistance of the cross section at a sufficient number of sections of the structure for the associated plastic hinges to constitute a mechanism involving the whole structure or some part of it

• Yield condition: At every cross section of the structure, the bending moment must be less than or equal to, the full plastic moment of resistance.

Advantage of plastic theory

• It is not necessary to consider continuity

• This makes plastic analysis considerably easier than the elastic analysis of any given structure

Plastic theory

• In many structures, there a number of alternative collapse mechanisms

• The correct mechanism is not immediately obvious

• It is necessary to approach correct solution in a series of steps

• The following theorems help in arriving at an acceptable solution

• No proofs are given as proofs are complicated

Method 1:Upper Bound Theorem

• Also called Kinematic theorem or Minimum Principle • In the analysis of a structure, an arbitrary choice of

collapse mechanism will lead to an estimate of the collapse load which is greater than or equal to the correct one

• If the correct collapse mechanism is not known and a guess is made, then the solution will give an upper bound on collapse load factorλc and is potentially unsafe.

• Methods based on assumed collapse mechanism generally satisfy only the equilibrium and mechanism condition

Method 2: Lower bound theorem

• Or Static theorem or maximum Principle

• An arbitrary equilibrium condition which also satisfies the yield condition will lead to an estimate of the collapse load which is less than or equal to the correct one

• Satisfying the equilibrium and yield condition without necessarily obtaining a mechanism is essentially safe procedure

Uniqueness theorem

• The value of the collapse load which satisfies the three conditions of equilibrium, mechanism and yield is unique.

Applications

• Fixed beam subject to two concentrated load

• Continuous beams

• Portal frames

Fixed beam subject to two concentrated load

• The beam shown in figure is fixed at both ends. It is subject to loads 20kN and 30 kN. If the full plastic moment capacity Mp is 78.0kNm, then determine the load factor against collapse.

• Answer:

• Method 1: Steps:

• (1) Consider all possible collapse mechanisms

4 m 4 m 2 m

20kN 30 kN

4 m 4 m 2 m

20kN 30 kN

20kN 30 kN

20kN 30 kN

20kN 30 kN

Equilibrium between bending moments and applied loads is satisfied by using the virtual work method

20λ 30λ

78 78

78

θ φ

4 m 6 m

First collapse mechanism

• The method considers a virtual displacement of the mechanism that is under investigation assuming that all internal strain is concentrated in the plastic hinges

• The members between the plastic hinges are assumed to be perfectly rigid and make no contribution to the internal work

• For compatibility 4θ = 6φ • By principle of virtual work, work done by

external load = work done internally • 20λx 4θ + 30λ x 4φ =78(2θ+2φ) • 20λx 6φ + 30λ x 4φ =78(3φ+2φ) • λ=390/240=1.625

Second collapse mechanism 20kN 30 kN

78

78

78

θ φ

For compatibility 4θ = 6φ By principle of virtual work, work done by external load = work done internally 30λx 4θ + 20λ x 4φ =78(2θ+2φ) 30λx 6φ + 20λ x 4φ =78(3φ+2φ) λ=390/260=1.5 All possible collapse mechanisms have been considered. Using minimum principle, we can confidently choose the mechanism with the lowest load factor. Therefore λ=1.5

6 m

4 m

Method 2

• Method 2 combines free and reactant bending moment diagram to obtain solution that satisfies the equilibrium and yield condition.

Continuous beams

• One of the advantages of plastic theory is that there are no compatibility conditions in the analysis

• As plastic hinges form, they destroy the continuity of the deflection profile and an important consequence of this in continuous beams is that the collapse of any span is independent of the adjacent spans

Continuous beams

• Consider a uniform beam of several equal spans.

• Let w be the unfactored load acting on it.

• It is required to find the load factor λ at collapse.

• There are only 2 collapse cases to consider namely midspan and endspan.

Internal span • Each of the internal span has exactly the same collapse

mechanism.

L/2 L/2

Mp

Mp Mp

θ θ

Udl

Factored load =λw

For the interior span using the work

equation method (free and reactant

bending moments is equally simple)

λwL x Lθ/4 = Mp(θ+2θ+θ)

Mp = λwL 2 /16

λ =16Mp/ wL 2

The left side of work equation

represents the virtual work done by

the udl. Here a virtual mechanism is

considered in which the members

remain perfectly straight between

plastic hinges. The distributed load

can therefore be considered to be

concentrated at the centroids of the

straight lengths of the members.

Derivation of external work term

L/4 L/4

θ θ

λwL/2 λwL/2

Lθ/4

L/4 L/4

External span

• The collapse condition in the end span is not so obvious because it is not clear where the internal plastic hinge is located.

• The kinematic theorem implies that all possible locations of the hinge must be considered.

• Choose one which gives the lowest load factor (or the highest value of Mp in a design calculation).

• This can be conveniently done by choosing an arbitrary hinge position defined by variable x, and using the method of calculus.

• The rotations θ and φ are related by considering the vertical deflection at the hinge position

• xθ = (L-x) φ

• The work equation can be written down as

x L-x

Mp

Mp Mp

θ φ

Udl

Factored load =λw

• λwL .Xθ/2=Mp(2θ+φ)=Mp(2+x/(L-x))θ

• I.e . Mp=λwL.x(L-x)/(2(2L-x)

• The critical value of x is that which maximises Mp i.e dMp/dx=0

• Solving, x=L(2±√2)

• This gives a unique root within the span 0 to L

• x=L(2-√2)=0.586L

• Backsubstituting

• Mp = λwL2/11.66 or λ=11.66Mp/wl2

• The end bay condition occurs frequently in practical design

• Note that if a beam is designed to be fully continuous over several spans, the end bays require a considerably stronger section than the internal bays (Mp = λwL2/11.66 compared to λ =16Mp/ wL 2 )

• The fabrication of connections that are adequate to ensure full continuity is expensive and a fully continuous uniform bean may well not be the optimum practical solution

Plastic analysis of portal frames

• The collapse mechanism in portal frames is not obvious

• Necessary to consider several possibilities

• Two types of portal frames

– Rectangular portal frames

– Pitched portal frames

RECTAGULAR PORTAL FRAMES

• The rectangular frame shown in figure has a uniform full plastic moment of 20knm. The loads shown are unfactored. Find the load factor λ against collapse. Use the work equation method

10Kn

5 M

5KN

3.75M 3.75M

A

B

C D

E

RECTAGULAR PORTAL FRAMES

• The only possible locations for plastic hinges are at A,B,C, D and E where there is a change of slope in the bending moment diagram

• It is impossible for plastic hinges to form between these points which are termed as CRITICAL SECTIONS

• Thus THERE ARE ONLY 3 VALID MECHANISMS shown below:

Different collapse mechanisms

10λ

θ

θ

5λ

Sway mechanism

10λ

θ θ

5λ

Beam mechanism

2θ

10λ

θ

θ

5λ

Combined mechanism

θ 2θ

2θ

• For a small, rigid-link movement of the mechanism, the beam moves bodily sideways and there is no downward movement of the 10kN load

• Work equation is 5λ x 5θ=20(θ+θ+θ+θ) • λ=80/25=3.2 • In plastic collapse mechanism, the bending moment at a

plastic hinge is always related to the direction of rotation of that hinge with the result that plastic hinges always do positive virtual work.

10λ

θ

θ

5λ

Sway mechanism

• For a small movement of the mechanism, the stanchion remain vertical and there is no movement of the 5 kN load. The work equation is therefore 10λx 3.75θ = 20(θ+2θ+θ)

• Therefore λ=80/37.5=2.13

10λ

θ θ

5λ

Beam mechanism

2θ

• Combined mechanism: this a combination of 1 and 2. It is essential to carry out this combination in such a way that the hinge at B is eliminated and replaced by a rigid joint. If this were not the case , it would be impossible to relate all movements in the virtual mechanism to a single variable θ

• 5λx 5θ + 10λx3.75θ=20(θ+2θ+2θ+θ)

• λ=120/62.5=1.92

10λ

θ

θ

5λ

Combined mechanism

θ 2θ

2θ

• All possible mechanisms have been considered. It follows from the minimum principle that the combined mechanism with the lowest load factor is the correct collapse mechanism.

• The load factor against collapse is 1.92

References

• Plastic design to BS 5950, Davies, J.M. and B.A.Brown, Blackwell Science, Steel Construction Institute, 1996.